Unit 1: Measurements and SI Units

Physics is the study of non-living and the way they work. Some of the topics covered are forces, energy, moments, light and electricity. In physics discoveries depend on making observations and doing experiments. When we make an observation such as” the metal bar expands when heated”; we are making a qualitative statement. When we are taking measurements and we state that “the metal bar expands 2.6 mm”; we are making a quantitative statement. All measurements have units and in order to avoid confusion, we use S.I. Units (a universal and standard system of measurement) International System of Units

Fundamental Quantity Base SI Unit

Name Symbol Name Symbol

Mass m Kilogram kg

Length l Metres m

Time t Seconds S

Current I amperes A

Temperature T/ϴ Kelvin

degrees Celsius

K

°C

Luminous Intensity I candela cd

Common Prefixes with SI Units

Prefix Multiple Symbol

giga 109 G mega 106 M kilo 103 k

milli 10-3 m micro 10-6 µ nano 10-9 n pico 10-12 p

Example: Convert a) 6000g to kg 6000 ÷ 1000 = 6kg b) 7m to mm 7 × 1000 = 7000 mm

Standard Form

To write a number in standard form we try to get it in the following format. A × 10B

Where A is a number between 1 and 10 and B is either a positive or negative whole number. Examples: Convert the following into standard form a) 5700 c) 0.00729 1.57 × 104 7.29 × 10-3

b) 200 d) 0.000059 2.00 × 102 5.9 × 10-5

1) (7.5 × 103) + (1.4 × 105)

2) (1.8 × 104) × (1.2 × 105)

3) (9.6 × 102) ÷ (3 × 10-3)

Density, Mass and Volume

Definitions

Mass (m):

the mass of an object is the amount of matter it contains. (SI units: kg)

Volume (v):

the volume of an object is the amount of space the object takes up. (SI units: m3)

Density (ρ):

the density of an object is defined as it mass per unit volume.

Formula:

density = mass volume m p = m v p v Units: kg/m3 or g/cm3

In order to determine the density of material, we need to find the mass can easily determined accurately by using the scale. If the object is a regular shape object we can determine the volume by using the appropriate formula. However, if the object has an irregular shape we can determine its volume by measuring the amount of water displaced in a measuring cylinder.

Volume of Regular Shapes

Cylinder

h Volume = ∏ r2h Cube/ Cuboids h Volume = length × width × height V = lwh

w Cone

Volume = 1 ∏ r2h h 3

Sphere

Volume = 4 ∏ r2

3 r Prism h b Volume of prism = cross sectional area of triangle × length Where area of triangle = 1 × base × height 2 = 1 bh 2

Volume of Liquids

To find the volume of a liquid, we can either use a measuring cylinder, burette or pipette. When reading the volume from these instruments; you should read the bottom of the curved meniscus and make sure your eye is on the same horizontal as the meniscus. If you are using a measuring cylinder, it should be on a flat level surface.

Examples of Density

1) A piece of metal has a mass of 140g and a volume of 20cm3. Calculate the density of the metal. m = 140g v =20cm3 p = m v = 140 = 7g/cm3

20 2) A body has a density of 0.25g/cm3. If the mass is 120g, calculate the body’s volume. p = 0.25g/cm3 m = 120g v = m p

= 120 0.25 = 480 cm3

3) Calculate the mass of a sold gold coin of volume 2.1cm3, given that the density of the gold coin is 19g/ cm3. p = 19g/cm3 v = 2.1cm3

m = 19g/cm3 × 2.1cm3

=39.9g 4) The density of a container is 780 kg/m3, if the mass of the container is 15600g, calculate the volume of the container. p = 780 kg/m3 m = 15600g v = m = 15.6 p 780

GRAPH WORK Example 1 A student adds a substance to a measuring cylinder and recorded the mass and the volume as shown in the table below. Mass / (g)

5 15 25 30 60

Volume / (cm3)

1 3 5 6 12

(a) Plot a graph of mass (y - axis) against volume (x - axis) Using a scale of: y – axis: 2 cm represents 5 g

x – axis: 1cm represents 1 cm3

(b) Determine the gradient and state the units of the gradient.

(c) What does the gradient represent?

Example 2 Draw a graph using the information shown in the table below Mass / (kg)

1 2 5 6 12

Density / (kg/m3)

10 20 50 60 120

(a) Plot a graph of mass (y - axis) against density (x - axis) Using a scale of: y – axis: 2 cm represents 1 kg

x – axis: 1cm represents 10 kg/m3

(b) Determine the gradient and state the units of the gradient.

(c) What does the gradient represent?

BEST FIT LINES Sometimes when we plot our points we realise that they do not “line up” to give us the perfect straight line graph. We need to draw a “best fit line”. First, we try to balance the ruler so that it passes through as many points as possible. However, we need to make sure that

(i) the same number of points are above the “best fit” line as below the line. (ii) The points not on the line are the same distances from the line.

[Remember to ALWAYS choose new points which lie on the line to calculate the gradient] Example 1 Draw a graph using the information shown in the table below Mass / (g)

5 13 17 21 25 29 40

Volume / (cm3)

0 10 15 20 25 30 40

(a) Plot a graph of mass (y - axis) against volume (x - axis) Using a scale of: y – axis: 1 cm represents 2 g

x – axis: 2cm represents 5cm3

(b) Determine the gradient and state the units of the gradient.

Example 2 The force acting on different masses was recorded as shown below. Force / (N)

0 5 10 16 20 24

mass / (kg)

0 1 2 3 4 5

(a) Plot a graph of Force (y - axis) against mass (x - axis) Using a scale of: y – axis: 1 cm represents 1 N

x – axis: 2cm represents 1 kg

(b) Determine the gradient and state the units of the gradient.

(c) If an object made of the same material used in the experiment has a mass of 3.5 kg, what is the force acting on it?

Example 2 Draw a graph using the information shown in the table below Mass / (kg)

1 2 5 6 12

Density / (kg/m3)

10 20 50 60 120

(d) Plot a graph of mass (y - axis) against density (x - axis) Using a scale of: y – axis: 2 cm represents 1 kg

x – axis: 1cm represents 10 kg/m3

(e) Determine the gradient and state the units of the gradient.

(f) What does the gradient represent?

FORCES AND MOTION Force A force is either push or pull. It is a vector quantity, that means it has magnitude and direction. Force is measured in a unit called Newtons (N). Some examples of forces are:

a) Weight- downward pull of gravity on a body. b) Tension- the force in a stretched rope or string. c) Friction- a force which stops objects. d) Air/Water Resistance- examples of friction.

Combining Forces (Resultant Force) On Earth, very few objects have just one force acting on them. Usually there at least two forces acting though the same point. When this occurs we can combine the forces to form the Resultant Force. Examples:

a. 3N 5N 2N Resultant Force (5-3) = 2N b. . 2N 5N 7N Resultant Force (2+5) = 7N c. 4N 4N

Resultant Force Zero (0N) Force, Mass and Acceleration We can state a relationship which links force, mass and acceleration together. This relationship is known as newtons second law in motion. Force = mass × acceleration F = ma F m a Definitions: Mass- the mass of an object is the amount of matter it contains. The mass is unchanged no matter where you are in the universe. (mass is measured in kg). Acceleration- Acceleration is the change of speed with time and it is measured m/s2 or ms2. Acceleration = force mass a = f m v a c Weight- Weight is the pull of a body caused by gravity. The weight of an object depends on where the object is in the universe. We can determine the weight by using the formula. Weight = mass × gravity. (since weight is a force its units is Newton) w m g

Mass- 100kg mass-100kg g=10m/s g = 1.3m/s 10N/kg = 1.3N/g Earth Moon w = mg w = mg = 100 × 10 = 100 × 1.3 = 1000N = 130N Weight on earth = 1000N Weight on moon = 130N On Earth a man of mass 100kg weighs 1000N. Since gravity on Earth is 10m/s2

or 10 N/kg.

On the moon the same man of mass 100kg weighs 130 N/kg. Since gravity on the moon is 1.3m/s2 or 1.3N/kg.

Balanced Forces

Sir Isaac Newton was the first to describe how objects would move if no forces were acting on them. His first law of motion states; If an object has no force on it, it will remain stationary, if was still.

OR

If it was moving, it will continue moving at a steady speed in a straight line. Friction

On Earth vehicles quickly come to rest because they are slowed by the force of friction. Friction is the resistance which must be overcome when one surface moves over the other. Friction always acts in the opposite direction to the movement and always opposes you when you try to do work. Although, surfaces may appear to be smooth they are in fact very rough. When an attempt is made to move two surfaces the hills and valleys interlock with each other. These interlocking cause friction between the two surfaces. When objects slide across each other like this the friction heats them up. It also causes the surfaces to wear away. Fluid Friction Liquid and gases are called fluids and they can also cause friction. Example: When a car is travelling down the highway it experiences a resistance which increases with a speed (more speed, more air resistance). Forces In-Balance

Most objects have several forces acting on them. Sometimes all the forces cancel out each other and the object behaves as if no force was acting on it at all. Action and Reaction

No force exists by itself, forces always occur in pairs. One force acts on the object while the other is equal but is opposite to its partner. This leads us into Newtons third law of motion; For every action there is an equal but opposite reaction.

OR

When Body A pushes on Body B then Body B pushes back on Body A, with an equal but opposite force.

Examples: of pairs of forces.

As a skydiver freefalls the air resistance increases as the speed increases. Eventually the air resistance is going to stop accelerating and falls at a maximum speed. This speed is called Terminal Speed which is approximately two metres per second.

Energy, Work and Power

Work is done whenever an object is moved by a force. We can measure the amount of work done by using the formula.

Work done = force × distance W = f × d (where the distance is in the direction of the force)

W

f d

The unit of work is Joule (J). Therefore 1J of work done when a force of 1N makes an object a distance of 1m. ( in the same direction of the force)

NB: 1J = 1Nm

Examples of doing work

a) If you lift an object vertically through a distance you have done work However if you stand still while holding the same object even though there are forces acting there is no movement. You will get tired after a while but you have done no work.

b) If you slide a box across the floor you are moving it against friction, hence you are doing work, however if you push against a wall, the wall would exert a resistance but there will be no movement hence no work is done.

1) Calculate the work done when a mass of 5kg is lifted 50cm.

2) Calculate the work done when a mass of 300g is lifted 150cm.

3) A lift of mass 250kg carries a man of mass 76kg, how much work is done by the wind rope which draws the lift through a height of 15m

4) A wind pump raises 2500N of water to the surface from 6m below the ground level. Find the work done by the pump.

Energy

In order to do work we must have a source of energy (energy to the capacity for doing work). They are many different forms of energy.

1) Heat / Thermal 2) Chemical 3) Potential 4) Kinetic 5) Light 6) Sound 7) Mechanical 8) Magnetic 9) Electrical 10) Nuclear

Devices which convert one form of energy into another are called energy converters or transducers. When energy is converted none is lost or gain even though some can be wasted (converted to a form which is not useful to us).

This leads us to the law of the conservation of energy.

The law states that energy cannot be created or destroyed but it can change from one form into another. For example:

Car converts chemical energy in the petrol into mechanical energy to move the car. However over 70% of this energy converted is wasted as heat energy in the radiator and the exhaust. More energy is also lost due to friction in various moving parts of the car. In general only 12% of the original energy is converted to move the car.

When an object is lifted against gravity work must be done. This work is stored in the lifted object. The higher the object is from its original position, the more energy it possesses. When this object is then released the potential energy (stored energy) is converted to kinetic energy (moving energy).

Object Mass in kg Height above ground

Definition

Potential Energy (PE) The potential energy of the body is the energy it possesses because of its position or state. We can calculate potential energy by using the formula: PE = mass × gravity (gravitational field strength) × vertical height PE = mgh Where m = kg g = N/kg h = m Units of PE = Joules (J) Example: A stone of mass of 200g is lifted 5m above the ground. Calculate its potential energy given that gravity is equal to 10 N/kg m = 200g h = 5m g =10 N/kg = 0.2kg PE = mgh = 0.2 × 5 × 10 = 10 J

Kinetic Energy (KE)

Kinetic Energy is the energy a body possesses because of its motion. KE = 1 mass × velocity squared 2 = 1 mv2

2 Where m = kg V = m/s Example:

A rock of mass 3kg has a speed of 6metres per second. Calculate its KE KE = 1 mass × velocity squared 2 = 1 × 3 × 62

2 = 1 × 3 × 36 2 =54J There are many examples of machines which convert potential energy into kinetic energy.

a) The stretched string of a bow possesses PE which is converted into KE of the moving arrow.

b) The water at the top of a waterfall possesses both PE and KE when it reaches the bottom all of this energy is converted to KE and it can be used to move turbines in electrical stages. If these machines are frictionless, then the PE and the KE are in a constant ratio. We can use the formula below to do calculations. PE lost = KE gained mgh = 1 mv2

2

Example:

A stone has a mass of 5kg is dropped 80m given that gravity is equal to 10N/kg. Calculate: a) PE of the stone

b) The velocity at which the stone will hit the ground 5kg

80m a) PE = mgh = 5 × 80 × 10 = 4000J b) When the stone strikes the ground the potential energy was converted to KE PE loss = KE gained 4000 = 1 mv2

2 4000 = 1 × 5 × v2 2 4000 = 2.5 × v2 v2 = 4000 2.5 v2 = 1600 v = √1600 v = 40m/s

Homework

NCSP3 Page 56 -57 # 1-2 (complete in test book for Friday)

Remember to complete pages 52 – 53 & pages 54 - 55

Power

When we speak about power we mean how quickly work is done. Power is the rate of doing work. Since work done is equal to energy transferred. We can also define power as the rate energy is transferred. We can calculate power by using the formula:

Power = Work done time

P = W t W/E OR Power = Energy time P t P = E t Where work done/ Energy is in Joules (J) time (t) is in seconds Units of power = Watts (W) 1 Watt is the power generated when 1 joule of work is done in 1 second. Example:

A forklift truck can lift a load 315kg to a height of 2m in 20 seconds. Calculate (a) the work done (b) the power of the forklift Mass = 315kg Distance = 2m Time = 20s a) Work done = force × distance Weight of load = mass × gravity = 315 × 10 = 3150N b) Power = Energy time = 6300 = 315W 20

v = 40 m/s Machine

A machine is a device which converts energy from one form to another. A force can be applied at one point and it is used to overcome a force at another point. The force being applied is called effort (E) and the force being overcome is called load (L). A machine in which the effort needed is less than the load is called a force multiplier. A machine in which the distance moved by the load is much greater than the distance moved by the effort is called a distance multiplier. Efficiency, Mechanical Advantage and Velocity Ratio

Mechanical Advantage (MA) = load effort

L

MA E

Velocity Ratio (VR) = distance moved by effort distance moved by load Efficiency (E) = useful work done × 100 total work done OR work got out × 100 work put in OR MA × 100 OR useful power output VR power input work done by load work done by effort

Efficiency is ideally 100% but all machines have an efficiency less than 100% because energy is lost. Levers

A lever is a simple machine which uses a pivot or a fulcrum to transfer the work done by the effort to a load. There are three types of levers. Your pivot is between the load and the effort. The lever works on the principle of moments; that means that the force applied to the lever turns it about the pivot. The turning effect depends on the size of the force used and how for away from the pivot it was applied. Moment of force = force × perpendicular distance moment = F × d Units = Nm Moments can either be clockwise or anticlockwise depending on the way they turn. e.g. A spanner produces a turning effect when we try to loose a nut.

We can increase the turning effects of the spanner by a) increasing the force applied or b) increasing the length of the spanner. e.g. Calculate the moment the force below.

Moments in Balance

Example: The See-Saw There are two turning effects on the see-saw a) Body A produces an anticlockwise turning effect on the se-saw. Ma = FA × dA b) Body B produces a clockwise turning effect on the see-saw. MB = FB × dB Since the see-saw is balanced the two turning forces are equal. We can therefore say that anticlockwise moments = clockwise moments {FA × dA = FB × dB} Example 1: A boy weighing 6000N sits at 6m from the pivot of a see-saw. A girl sits on the opposite side 9m from the pivot and balances the see-saw. How much does the girl weigh? Anticlockwise moments = Clockwise moments WB × dB = Wg × dg

600 × 6 = Wg × 9

3600 = 9Wg

Wg = 3600 9

Wg = 400N

Example 2: A boy of 80kg sits to the left of a plank 1m away from the pivot. A girl of mass 30kg sits 3m behind him. A man 100kg sits on the right side of the pivot and balances the plank. How far from the pivot is the man sitting? Clockwise moments = anticlockwise moments 1000 × dm = (800 × 1) + (300 × 4) 1000dm = 800 + 1200 1000dm = 2000 dm = 2000 1000

= 2m (the man in 2m from the pivot)